Limit

[BigBeachBanana]

I took log with the base of 10 instead of ln

I've edited Skeeter's base e to base 10.
$\displaystyle \log_{10}{y} = \lim_{x \to 1} \dfrac{\log_{10}(1+\sin(\pi x))}{\tan(\pi x)}$

L’Hopital …

$\displaystyle \log_{10}{y} = \lim_{x \to 1} \dfrac{\frac{\pi \cos(\pi x)}{\log(10)(1+\sin(\pi x))}}{\pi \sec^2(\pi x)}$

$\displaystyle \log_{10}{y} = \lim_{x \to 1} \dfrac{\cos^3(\pi x)}{\log(10)(1+\sin(\pi x))}= \frac{-1}{\log(10)}$

$y=10^{\frac{-1}{\log(10)}}=\frac{1}{e}$

 

[Loki123]

I've edited Skeeter's base e to base 10.
$\displaystyle \log_{10}{y} = \lim_{x \to 1} \dfrac{\log_{10}(1+\sin(\pi x))}{\tan(\pi x)}$

L’Hopital …

$\displaystyle \log_{10}{y} = \lim_{x \to 1} \dfrac{\frac{\pi \cos(\pi x)}{\log(10)(1+\sin(\pi x))}}{\pi \sec^2(\pi x)}$

$\displaystyle \log_{10}{y} = \lim_{x \to 1} \dfrac{\cos^3(\pi x)}{\log(10)(1+\sin(\pi x))}= \frac{-1}{\log(10)}$

$y=10^{\frac{-1}{\log(10)}}=\frac{1}{e}$

How did you get log after using L'Hopital. Doesn't it turn to xlna

 

[BigBeachBanana]

How did you get log after using L'Hopital. Doesn't it turn to xlna

They are. Notice I didn't write $\log_{10}(10)$, but simply $\log(10)$ which meant to be understood as $\log_{e}(10)=\ln(10)$

 

[Loki123]

10log(10)−1=e1

Could you write this process a bit more detailed because I don't get it.

 

[BigBeachBanana]

Could you write this process a bit more detailed because I don't get it.

$$10^{\frac{-1}{\ln 10}}=10^{\frac{-\ln e}{\ln 10}}=10^{-\log_{10}e}=e^{-1}$$Applying the change of base for log functions: $$\log_{a}b=\frac{\log_{c}a}{\log_{c}b}$$